Question

In answering a question on a multiple choice test a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that a student knows the answer given that he answered it correctly?

Answer 1

Let A, E₁ and E₂ denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{3}{4} \\ P\left(E_2\right)=\frac{1}{4} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=1 \\ P\left(A/E_2\right)=\frac{1}{4} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{3}{4}}\times {\displaystyle 1}}{{\displaystyle \frac{3}{4}}\times 1+{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{1}{4}}} \\ =\frac{3}{3+{\displaystyle \frac{1}{4}}}=\frac{12}{13} \end{gathered}$$